How to represent that there only only four unique possible prices in terms of probability (nCm)? Assume that in each period the stock can go either up by $20\%$ or down by $10\%$. What are the possible prices of the stock after three periods if its initial price is $S(0)=\$100$?
I know what the possible prices are, there are 8 different scenarios but only four of the scenarios are unique while the other four are duplicates, just in a different order.   
For example these two scenarios will gove you the same price, scenario 1: the stock goes up, down, then down and scenario 2: the stock goes down, up, then down.
Both of these scenarios will give you the same price because the order doesn't matter when it comes to whether the stock goes up or down, as long as it goes up or down in the same number of times.  
I forgot how to represent that there only only four unique possible prices in terms of probability (nCm)?
 A: I think what you are asking is what are the probabilities to get certain stock price after three periods.
I think the probabilities for the two extreme values(up&up&up  or down&down&down) are 1/8.
The two other middle values are of probabilities 3/8.
A: I guess since the order of ups and downs doesn't matter, you only have to count the number of ups possible in the three periods.
For example


*

*1 way to have 0 "up" in the three periods

*1 way to have 1 "up's" in the three periods (since order doesn't matter)

*1 way to have 2 "up's" in the three periods (since order doesn't matter)

*1 way to have 3 "up's" in the three periods


So there's 4 ways in total.
It's really not necessary to use (nCm) to calculate the result, but if you really want to, that would be
$\frac{3\choose 0}{3\choose 0}+\frac{3\choose 1}{3\choose 1}+\frac{3\choose 2}{3\choose 2}+\frac{3\choose 3}{3\choose 3}$
You need to divide each result ($\frac{3\choose x}{3\choose x}$) by number of orderings ($\frac{3\choose 0}{3\choose 0}$).
A: Binomial?
$$100(1.2)^u(0.9)^{3-u}$$
where $u= 0,1,2,3$
